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1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S, (s) + H₂O (1)→ Al(OH), (aq) +_____ H₂S (g) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams?

1. 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: _Al₂S, (s) + H₂O (1)→ Al(OH), (aq) +_____ H₂S (g) a. Which substance is the limiting reactant? b. How many liters of H₂S is formed? c. What amount of excess reactant is left over, in grams?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**How to Determine LR:** 1. Convert reactant quantities to moles. 2. Divide each mole amount by the balanced coefficient. 3. The smallest value is the limiting reactant (LR). **Additional Information:** - **Avogadro’s Number**: \(6.023 \times 10^{23}\) molecules/1 mole - **STP Volume**: 1 mole = 22.4 L - **Left Over**: Initial amount - Consumed amount --- ### Reaction Problem 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: \[ \mathrm{Al_2S_3(s) + H_2O(l) \rightarrow Al(OH)_3(aq) + H_2S(g)} \] **Questions:** **a. Which substance is the limiting reactant?** **b. How many liters of H₂S are formed?** **c. What amount of excess reactant is left over, in grams?** --- **Graph/Diagram Explanation**: In the given image, there are no graphs or diagrams. The text consists of instructions and a chemical reaction problem meant to be solved by students. Note to students: - To solve for the limiting reactant, follow the three steps outlined above. - Use Avogadro's number and STP volume as needed in your calculations. - Make sure to balance the chemical equation before solving the problem.
Transcribed Image Text:**How to Determine LR:** 1. Convert reactant quantities to moles. 2. Divide each mole amount by the balanced coefficient. 3. The smallest value is the limiting reactant (LR). **Additional Information:** - **Avogadro’s Number**: \(6.023 \times 10^{23}\) molecules/1 mole - **STP Volume**: 1 mole = 22.4 L - **Left Over**: Initial amount - Consumed amount --- ### Reaction Problem 15.6 g of aluminum sulfide reacts with 10.0 g of water according to the reaction below: \[ \mathrm{Al_2S_3(s) + H_2O(l) \rightarrow Al(OH)_3(aq) + H_2S(g)} \] **Questions:** **a. Which substance is the limiting reactant?** **b. How many liters of H₂S are formed?** **c. What amount of excess reactant is left over, in grams?** --- **Graph/Diagram Explanation**: In the given image, there are no graphs or diagrams. The text consists of instructions and a chemical reaction problem meant to be solved by students. Note to students: - To solve for the limiting reactant, follow the three steps outlined above. - Use Avogadro's number and STP volume as needed in your calculations. - Make sure to balance the chemical equation before solving the problem.
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